5

Let $B_n\subset\mathrm{GL}_n(\Bbb C)$ be the group of invertible upper-triangular matrices. What is the index $[\mathrm{GL}_n(\Bbb C):B_n]?$ (By index I mean the cardinality of a coset space.)

In this answer, it is shown that the index is infinite. It also follows from the conjunction of this and this answer. But which cardinal is it? I'm almost certain that it's $\mathfrak c,$ but I have no idea how to prove it.

More generally, which cardinals are the indexes $[\mathrm{GL}_n(\Bbb C):G],$ where $G$ runs over the set of subgroups of $\mathrm{GL}_n(\Bbb C)?$ According to the latter two linked answers, there shouldn't be any finite cardinals there. I suspect all of these cardinals are actually $\mathfrak c$, but I wouldn't know how to prove it either (and this suspicion is much weaker than the previous one).

Community
  • 1
Bartek
  • 6,265
  • For the index of Borel subgroup, it will be finite. You may like to consider the Bruhat decomposition: http://groupprops.subwiki.org/wiki/Bruhat_decomposition_theorem – Easy Mar 30 '13 at 13:55
  • @Easy I don't understand. How can it have a finite index? Doesn't it contradict the proofs in the linked answers? – Bartek Mar 30 '13 at 14:15
  • I think Easy is interpreting "index" as "dimension of the quotient." – Jason DeVito - on hiatus Mar 30 '13 at 14:19
  • @JasonDeVito What is dimension here? The groups involved don't seem to have any natural vector space structure, do they? Do you mean dimension of the manifolds? That's definitely not what I'm asking about. My question is purely about the algebraic structure. – Bartek Mar 30 '13 at 14:21
  • The groups are smooth manifolds (and the notion of "dimension" applies to these). – Jason DeVito - on hiatus Mar 30 '13 at 14:23
  • @Jason OK, I had no idea my question was ambiguous. I mean the cardinality of the quotient, not its dimension. I'll edit the question to make it clear. – Bartek Mar 30 '13 at 14:25
  • @Bartek: Well, index is well defined, so in that sense, there was no ambiguity. But to those of us who first consider $Gl(n,\mathbb{C})$ as a Lie group, cardinality rarely comes up, so we subconsciously asssociate "index" to "measure of size of quotient". :-) – Jason DeVito - on hiatus Mar 30 '13 at 14:29
  • I think Easy rather meant that the set of right and left orbits $ B\backslash GL_n(\mathbb{C}) / B$ is finite, which is different from $B\backslash GL_n(\mathbb{C}) \simeq GL_n(\mathbb{C}) / B$ – user10676 Mar 30 '13 at 14:29
  • @JasonDeVito Couldn't some facts about manifolds give a solution to my question? I know almost nothing about manifolds and Lie groups, but if the coset space is naturally a manifold, then wouldn't counting its connected components be enough? (Or rather counting the one-point components and the others separately.) I think a connected manifold with more than one point should have $\mathfrak c$ points, shouldn't it? – Bartek Mar 30 '13 at 14:38

1 Answers1

7

The group $GL_n(\mathbb{C})$ is divisible, because every invertible matrix $A$ can be written $A = \exp(B) = (\exp(\tfrac{1}{n}B))^n$. Hence every quotient of $GL_n(\mathbb{C})$ is divisible. In particular every finite quotient is trivial.

There does exist a countable quotient of $GL_n(\mathbb{C})$, because $\mathbb{Q}$ is a quotient of $\mathbb{R}$ (use axiom of choice to take $\mathbb{Q}$-basis of $\mathbb{R}$), and $\mathbb{R} \simeq \mathbb{R}_{>0}$ is a quotient of $\mathbb{C}^\times$, and the latter is a quotient of $GL_n(\mathbb{C})$.

EDIT : take $\Gamma$ a $\mathbb{Q}$-subvector space of $\mathbb{R}$ of codimension $1$ (hence $\mathbb{R}/\Gamma \simeq \mathbb{Q}$). Let $H$ be the subgroup $GL_n(\mathbb{C})$ consisting of matrices $A$ such that $\log(|\det A|) \in \Gamma$. Then the map $$GL_n(\mathbb{C})/H \longrightarrow \mathbb{R}/\Gamma, \quad (A \mod H) \mapsto (\log(|\det A|) \mod \Gamma)$$ is an isomorphism.

user10676
  • 8,521